Name the proces by which heat flows in solids
1 answer:
You might be interested in
<h2>Answer:</h2>
(a) Attached to the response as Figure 1.
(b) 35.0Ω
(c) Across 5.0Ω = 1.3V
Across 10.0Ω = 2.6Ω
Across 20.0Ω = 5.2Ω
<h2>Explanation:</h2>(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.
(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e
R = R₁ + R₂ + R₃ -------------------(i)
Where;
R₁ = 5.0 Ω
R₂ = 10.0 Ω
R₃ = 20.0 Ω
<em>Substitute these values into equation (i) as follows;</em>
∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω
∴ R = 35.0 Ω
Therefore, the equivalent resistance is ∴ R = 35.0Ω
(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;
i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.
ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;
V = I R ------------------(ii)
Where;
V = Voltage supplied to the circuit = 9.0V
I = Current through the circuit
R = Resistance of the equivalent resistor = 35.0Ω
Substitute these values into equation (ii)
9.0 = I x 35.0
I =
I = 0.26A
This is also the current flowing through each of the resistors separately.
iii. Calculate the voltage drop across
1.<em> 5.0 Ω resistor</em>
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 5.0Ω resistor
I = current through the 5.0Ω resistor = 0.26A
R = resistance of the 5.0Ω resistor = 5.0Ω
=> V = 0.26 x 5.0
=> V = 1.3V
2.<em> 10.0 Ω resistor</em>
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 10.0Ω resistor
I = current through the 10.0Ω resistor = 0.26A
R = resistance of the 10.0Ω resistor = 10.0Ω
=> V = 0.26 x 10.0
=> V = 2.6V
3.<em> 20.0 Ω resistor</em>
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 20.0Ω resistor
I = current through the 20.0Ω resistor = 0.26A
R = resistance of the 20.0Ω resistor = 10.0Ω
=> V = 0.26 x 20.0
=> V = 5.2V
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m